The solution of the differential equation cos | vedclass

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(D) Given the differential equation: $\cos x \, dy = y(\sin x - y) \, dx$Divide both sides by $\cos x \, dx$:$\frac{dy}{dx} = y 	an x - y^2 \sec x$Re

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$\sec x = (	an x + c)y$

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(D) Given the differential equation: $\cos x \, dy = y(\sin x - y) \, dx$Divide both sides by $\cos x \, dx$:$\frac{dy}{dx} = y 	an x - y^2 \sec x$Rearrange the terms:$\frac{dy}{dx} - y 	an x = -y^2 \sec x$Divide by $y^2$:$y^{-2} \frac{dy}{dx} - y^{-1} 	an x = -\sec x \quad \dots(1)$Let $v = y^{-1} = \frac{1}{y}$. Then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,or $y^{-2} \frac{dy}{dx} = -\frac{dv}{dx}$.Substituting into equation $(1)$:$-\frac{dv}{dx} - v 	an x = -\sec x$$\frac{dv}{dx} + v 	an x = \sec x$This is a linear differential equation of the form $\frac{dv}{dx} + Pv = Q$,where $P = 	an x$ and $Q = \sec x$.The Integrating Factor ($I$.$F$.) is $e^{\int P \, dx} = e^{\int 	an x \, dx} = e^{\ln|\sec x|} = \sec x$.The general solution is $v \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c$.$v \sec x = \int \sec x \cdot \sec x \, dx + c$$v \sec x = \int \sec^2 x \, dx + c$$v \sec x = 	an x + c$Since $v = \frac{1}{y}$,we have:$\frac{1}{y} \sec x = 	an x + c$$\sec x = y(	an x + c)$

The solution of the differential equation $\cos x \, dy = y(\sin x - y) \, dx$ for $0 < x < \frac{\pi}{2}$ is:

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